class 10 “stability” - universidade da beira...

Class 10“Stability”

The system is stable if the output to the impulse input → 0 whenever t → ∞

outputinputS

whenever the input r(t) = impulse

That is, if the output of the system satisfies

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lim y(t)│t → ∞ = 0

outputinputS

There are other definitions which are equivalent, such as:

The system is stable if for every limited input there is a limited output

Due to this definition, stable systems are commonly called by BIBO-stable

(BIBO = bounded input-bounded output)

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A system is stable if, and only if, it has all its poles with negative real parts

That is, a system is stable if it has got allits poles located in the left half plane (LHP)

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outputinputS

So, the stability of systems can be determined by the location of the poles of the system in the complex plane

It is necessary that ALL the poles of the system lie in the LHPin order to be a stable system.

One single pole that do not lie in the LHP ruins the stability

turning it in a unstable system.

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Example 1:

Consider the de 1st order system

whose transfer function is given by:

)as(

1

)s(U

)s(Y

−=

The only pole of this system is located in

which can be in the LHP (left half plane), , on the imaginary axis or in the RLP (right half plane), depending on the value of a

(a < 0, a = 0 or a > 0, respectively)

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s = a

x’ = a x + u

y = x

thus,

Example 1 (continued):

a < 0

a = 0

a > 0

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Case a < 0, system is stable

output to the unit impulse output to the unit step

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Case a = 0, system is neither stable nor unstable


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Case a > 0, system is unstable

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summarizing, Example 1 (continued):

stable

indifferent

unstable

a < 0

a = 0

a > 0

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Example 2:

Consider the 2nd order system with transfer function given by

4s2s

4

)s(U

)s(Y2 +−

=

this system has got a pair

of complex poles with

positive real parts

s = 1 ± j·1,732

poles in the RHP

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Complex plane


unstable system

output to the unit step input

ζ = – 0,5ω = 2

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Example 3:

Consider the 2nd order system with transfer function given by

4s2s

4

)s(U

)s(Y2 ++

=

this system has got a pair of complex poles with negative real parts

s = –1 ± j·1,732

poles in the LHP

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Complex plane

stable system

output to the unit step input


ζ = 0,5ω = 2

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Summarizing: the stability of systems depend only on the location of its poles, which have all to be in the LHP in order to the system to be stable

Thus, the stability do not depend of the whole transfer function, but only on its denominator, the

characteristic polynomial p(s) of the system

that gives us the poles of the system

With respect to the state equation

x = A x + B u

y = C x + D u

⋅

the stability do not depend on B, C or D, but only on the matrix Athat gives us the characteristic polynomial and the poles of the system

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Routh-Hurwitz stability criterion

Routh, in the XIX century yet, in a time that there was no electricity, calculating machine

or computer, create a mathematical method in order to determine the number of poles of a system located in the RHP without having to calculate the poles themselves

(canadiano, 1831-1907)

Edward John Routh

This made easier the determination of the stability of a system

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Hurwitz is another important mathematician in the field of control and dynamic systems

Hurwitz derived in 1895 what is called today the

“Routh-Hurwitz stability criterion”

for determining whether a system is stable

The Routh-Hurwitz criterion is constructed from the characteristic polynomial of the system

n1n

2

2n

1n

1

n

o asasasasa)s(p +++++= −−−

L (ao > 0)

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It is said that Hurwitz did it independently of Routh who had derived it earlier by a different method

(German, 1859-1919)

Adolf Hurwitz

sn ao a2 a4 a6

sn-1 a1 a3 a5 a7

sn-2

sn-3

::s2

s1

so

Routh-Hurwitz table, initially fulfilled with the coefficients of

the characteristic polynomial p(s)

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from p(s) the Routh-Hurwitz table is built:

sn ao a2 a4 a6

sn-1 a1 a3 a5 a7

sn-2

sn-3

::s2

s1

so

If the characteristic polynomial p(s) is odd, the last element stays

in the 2nd line

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Routh-Hurwitz table, initial fulfilling

sn ao a2 a4 a6

sn-1 a1 a3 a5 0

sn-2

sn-3

::s2

s1

so

If the characteristic polynomial p(s) is even, the last element stays

in the 1st line, and a ‘0’ (zero) is put underneath in the 2nd line

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sn ao a2 a4 a6

sn-1 a1 a3 a5 a7

sn-2

sn-3

::s2

s1

so

After the initial fulfilling the remaining lines are calculated as will be indicated a further here

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sn ao a2 a4 a6

sn-1 a1 a3 a5 a7

sn-2 b1 b2 b3 b4

sn-3 c1 c2 c3 c4

: : : : :: : : :s2 d1 d2

s1 e1

so f1

Routh-Hurwitz table, after completed

After being completed the Routh-Hurwitz table

will have (n+1) lines

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sn ao a2 a4 a6

sn-1 a1 a3 a5 a7

sn-2 b1 b2 b3 b4

sn-3 c1 c2 c3 c4

: : : : :: : : :s2 d1 d2

s1 e1

so f1

Routh-Hurwitz table, after completed

Lines gets shorter and shorter

up to the 2 last lines

that have only one element each

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sn ao a2 a4 a6

sn-1 a1 a3 a5 a7

sn-2 b1 b2 b3 b4

sn-3 c1 c2 c3 c4

: : : : :: :s2 d1

s1 e1

so f1

Routh-Hurwitz table, pivot column

The first column (where are the

coefficients ao and a1) is called ‘pivot column’

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sn ao a2 a4 a6

sn-1 a1 a3 a5 a7

sn-2 b1 b2 b3 b4

sn-3 c1 c2 c3 c4

: : : : :: :s2 d1

s1 e1

so f1


The following elements of the

‘pivot column’ are: b1 , c1 , d1 ,

e1 , f1 , etc.

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sn ao a2 a4 a6

sn-1 a1 a3 a5 a7

sn-2 b1 b2 b3 b4

sn-3 c1 c2 c3 c4

: : : : :: :s2 d1

s1 e1

so f1


The elements of the ‘pivot column’

are called ‘pivots’

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The elements b1 , b2 , b3 , … etc. of third line

Note that in the calculations of the elements of this line there is

a division by a1, the pivot element of the previous line

1

3021

31

2o

1

1a

aaaa

aa

aadet

a

1b

⋅−⋅=

⋅−=

1

5041

51

4o

1

2a

aaaa

aa

aadet

a

1b

⋅−⋅=

⋅−=

1

7061

71

6o

1

3a

aaaa

aa

aadet

a

1b

⋅−⋅=

⋅−=

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The elements c1 , c2 , c3 , … etc. of third line

Note that in the calculations of the elements of this line there is

a division by a1, the pivot element of the previous line

1

2131

21

31

1

1b

baab

bb

aadet

b

1c

⋅−⋅=

⋅−=

1

3151

31

51

1

2b

baab

bb

aadet

b

1c

⋅−⋅=

⋅−=

1

4171

41

71

1

3b

baab

bb

aadet

b

1c

⋅−⋅=

⋅−=

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In the same way we can calculate the remaining elements of all lines and columns and the Routh-Hurwitz table is completed

Note that, as already mentioned earlier, the lines of the Routh-Hurwitz table gets shorter and shorter as there is less elements to be computed in the last positions in every line

Observation: Routh-Hurwitz table was built supposing that the

coefficient ao of the characteristic polynomial p(s) is positive,

ao > 0

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This happens since, as it is well known, the roots of a polynomial

do not alter when all its coefficients are multiplied by –1 (or by any

other constant value ≠ 0).

If the coefficient ao is negative,

ao < 0

then we can redefine p(s) with all its coefficients having the signs

changed (i.e., multiplied by –1)

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Therefore we get ao > 0

This will not alter the result to be obtained from the

Routh-Hurwitz table

The Routh-Hurwitz table allows us to find out how many poles are located in the RHP

The number of sign changes in the pivot column of Routh-Hurwitz table is the number of poles in the RHP

To simplify the calculations, if we wish we can multiply (or divide) all

elements of any line of the Routh-Hurwitz table by a positive

number

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sn ao a2 a4 a6

sn-1 a1 a3 a5 a7

sn-2 b1 b2 b3 b4

sn-3 c1 c2 c3 c4

: : : : :: :s2 d1

s1 e1

so f1

Routh-Hurwitz table

the number of sign changes in the pivot column is the number of poles in the SPD.

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The determination of the number of poles in the RHP can be very useful but, nevertheless, it doesn’t give us a diagnosis for stability of the system immediately.

That is because a system to be stable must have all its poles in theLHP and therefore, even if the number of sign changes in the pivot

column is zero, this only means that there will be ‘zero poles’ in the RHP, which does not guarantee stability yet, since it may have some pole in the imaginary axis

However, poles in the imaginary axis will reflect in zeros in the pivot column. These allows us to write the following result:

A system has all its poles in the LHP if, and only if, all the coefficients of the pivot column of the

Routh-Hurwitz table are positives

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sn ao a2 a4 a6

sn-1 a1 a3 a5 a7

sn-2 b1 b2 b3 b4

sn-3 c1 c2 c3 c4

: : : : :: :s2 d1

s1 e1

so f1

Routh-Hurwitz table

The system has all its poles in the LHP whenever all its pivot

column elements are positives

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1 3 5

2 4 0

1 5

-6

5

Consider now the characteristic polynomial of degree 4 given below

Example 4:

Setting up the Routh-Hurwitz table we get

2 sign changes (of

elements in the

pivot column)

Thus, the system is not stable

This characteristic

polynomial has 2 poles in the RLP

s4

s3

s2

s1

s0

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1 3 5

1 2 0

1 5

-3

5

2nd line ÷ 2


Here we could, for example, to simplify the calculation, divide the 2nd line (i.e., line s3 ) by 2This do not alter the final result

s4

s3

s2

s1

s0

and we get to the same conclusion:2 sign changes (of

elements of the

pivot column)

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Example 5:

Find the values of K for which the closed loop system

below is stable

Computing the closed loop transfer function (CLTF), we get

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1 2 K

2 2 0

1 K

(2-2K)

K


So, the characteristic polynomial of the closed loop system is given below

The Routh-Hurwitz table for this polynomial is:

s4

s3

s2

s1

s0

> 0

> 0

K < 1 0 < K < 1

system is

stable for

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Example 6:

Find the values of K for which the closed loop system below is stable

It is easy to verify that the characteristic polynomial of this system is:

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1 6 K

2 4 0

4 K

(4-K/2)

K


thus, the Routh-Hurwitz table for this polynomial is:

s4

s3

s2

s1

s0

> 0

> 0

K < 8 0 < K < 8

system is

stable for

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1 1

3 3

0

3

Example 7:

s3

s2

s1

s0

≅ ε > 0

so, all the elements of the pivot column arepositives, that is

0 (zero) roots in the RHP

Actually, this ‘zero’ in the pivot column indicates a symmetry of 2 poles and they can only be located in the imaginary axis

Consider now the characteristic polynomial p(s) of 3rd degree below

The Routh-Hurwitz table is also given below

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1 1

3 3

0

3

Example 7:

s3

s2

s1

s0

≅ ε > 0

so, all the elements of the pivot column arepositives, that is

0 (zero) roots in the RHP

Thus, the system is not unstable, but it does not have all its poles in the LHP either, that prevents stabilization

Consider now the characteristic polynomial p(s) of 3rd degree below

The Routh-Hurwitz table is also given below

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1 13 6

6 14 0

32/3 6

340/32 0

6

0 0 (zero) in pivot column

Example 8:

s5

s4

s3

s2

s1

s0

so, there is no sign change in the pivot

column, and that means

0 (zero) roots in RHP

Consider now the characteristic polynomial p(s) of 5th degree below

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That zero in the pivot column is in the last position. (It indicates symmetry of 1 pole with respect to the origin, that is, one pole = zero

1 13 6

6 14 0

32/3 6

340/32 0

6

0 0 (zero) in pivot column

Example 8:

s5

s4

s3

s2

s1

s0

So, the system is not unstable, nevertheless only 4 of its 5 poles are

in the LHP, since one is in the origin (s = 0)

so, there is no sign change in the pivot

column, and that means

0 (zero) roots in RHP

Consider now the characteristic polynomial p(s) of 5th degree below

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1 12 -29

3 36 -87

0 0 0 0 in pivot column

line of zeroes

Example 9:

s5

s4

s3

Consider now the characteristic polynomial below of 5th degree

and find the derivative q’(s)

from line s4 we get q(s)

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1 12 -29

3 36 -87

0 0 0

12 72 0

18 -87

130

-87

line of zeroes out

line q’(s) gets in

s5

s4

s3

s3

s2

s1

s0

By eliminating the line of zeroes, we can complete the Routh-Hurwitz table

1 sign change (of

elements of the

pivot column)

Thus, the system is not stable.

This characteristic

polynomial has 1 pole in RHP

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relative stability

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Relative Stability

Sometimes it is desirable that the poles do not lie close to the imaginary axis �� (in order to avoid having slow input responses

or with excessive oscillations)

What makes the stabilization to be quicker or slower is the localization of the poles to be further or closer (to the left) to the imaginary axis

The further to the left are the poles of the system, the quicker it stabilizes

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0

x

x

Re(z)

Im(z)

pólos afastados do

eixo imaginário

System Awith poles far from imaginary axis

System Bwith poles close to the imaginary axis

Consider these 2 systems:

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step response

takes only 25 seconds to stabilize

System A stabilizes quicker than system B since it has its poles

further to the imaginary axisSystem A

System Btakes 250 seconds to stabilize

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We have already seen that a system is stable or not ifit has all its poles in the LHP or not, respectively

That depends on how further away from the imaginary axis, to the left, are located the poles of this system

However, we have just seen too that: a stable system can be more stable than other

That is the concept of “relative stability”

That is the concept of “ABSOLUTE STABILITY”.

A localização dos polos é o que conta

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So, it is possible that we wish to verify if a system has its poles

lying to the left of the line s = −σ in the LHP,

and not only in the LHP

σ−

σ−=s

Region of the

LHP to the left

of the line s =

− σ

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p� s = p s − σ

p s To obtain that we do a translation

s ↦ s − σ

of the characteristic polynomial

p s so that we get

p� s = p s − 2

In this example

σ = 2

σ = 2

p� s = p s − σ

Which is the polynomial p s

shifted to � to the right

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p� s = p s − σp s

σ = 2

Translation of the polynomial to the right

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A translation of the polynomial to the right corresponds to

shift the line s = − σ to the right

or, equivalently,

shift the imaginary axis to the left

roots of p s roots of p� s = p s − σ

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Applying the “Routh-Hurwitz stability criterion” to the translated

polynomial p� s , we have that o number of sign changes in the

pivot column is equal to the number of roots of p s located to the

right of the line s = −σ

That is, through p� s we can extract conclusions

for p s

For example: If in Routh-Hurwitz table

for p� s there is no sign

change in the pivot column

(‘zero change’), then p swill not have poles

(‘zero poles’) to the right

of line s = −σ

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As p� s is a translation

of p s by σ units to the right, then all the conclusions extracted

for p� s with respect to the imaginary axis

are valid to p s with

respect to the line s =

− σ

In other words, the poles will be located in the region shown in the figure:

to the left of the line s = −σ, or in the line itself

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Verify if the characteristic polynomial p(s) given below has its

poles lying to the left of s = −2p s = s + 8s� + 21s + 20

Let us verify first if p s has all its roots in the LHP

p s = s + 8s� + 21s + 20

The pivot column of Routh-Hurwitz table of p s is all positive

and therefore, the polynomial has its poles in the LHP

1 21

8 20

18,5

20

zero changes in the pivot column

s3

s2

s1

s0

Example 10:

1 1

2 2

0

2

s3

s2

s1

s0

≅ ε > 0 0 (‘zero’) in the pivot column

But, will there be poles to the left of the line s = −2? In order to

find this we need to calculate the polynomial p� s with � = 2

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p� s = p s − 2 = s − 2 + 8 s − 2 � + 21 s − 2 + 20

= s + 2s� + s + 2

There is no sign change in the pivot column


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This means that there is no roots of p� s in the RLP and then, there is no roots of p s to the right of line s = −2

But that ‘zero’ in the second line from below (line ��) in the pivot

column indicates that there are 2 roots of p� s in the imaginary axis

and therefore, there are 2 roots of p s lying on the line s = −2

By verifying the location of the roots of p s we can ratify that:

o there is no pole to the right of the line s = −2

o there is 1 pole to the left of the line s = −2 (at s = −4)

and also that:

o there are 2 poles on the line s = −2

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The roots of p s are:

s = −4

s = −2 ± j

Location of the roots of

polynomial p s in the complex plane


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The 2 roots of p� s in the imaginary axis correspond to the 2 roots

of p s in line s = −2

roots of p s in

line s = −σ

roots of p� sin the imaginary axis

Location of the 2 roots of the

polynomials p� s and p s


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Verify if the characteristic polynomial p(s) given below has its poles

to the left of s = −1

Example 11:

p s = 2s� + 13s + 28s� + 23s + 6

Firstly let us verify if p s has all its roots located in the LHP

2 28 6

13 23 0

24,46 6

19,81 0

6

s4

s3

s2

s1

s0

the pivot column is all positive and therefore

this polynomial p s has all its poles lying in the SPE

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However, we want to know if they lye to the left of the line s = −1

For this we have to calculate the polynomial p� s by setting σ = 1

p� s = p s − 1 = 2 s − 1 � + 13 s − 1 + 28 s − 1 � + 23 s − 1 + 6

= 2s� + 5s + s� − 2s

2 1 0

5 − 2

9/5 0

− 2

0

s4

s3

s2

s1

s0

there is 1 sign change in the pivot column, thus

0 (‘zero’) in thepivot column

there is one root of p s to the

right of the line s = −1

there is one root of p� s in the RHP and therefore,

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Exemplo 11 (continuação):

But that ‘zero’ in the last line (line ��) of the pivot column

indicates that there is 1 root of p� s lying in the imaginary axis

(at the origin) and therefore, there is 1 root of p s on s = −1

By verifying the location of the roots of p s we can ratify that:

o in fact there is 1 pole lying on the line s = −1

and that

o There is also 1 pole lying to the right of the line s = −1 (at s =

− 0,5), which obviously is in the interval {−1 < s < 0} since we

have already seen that p s has no roots in the RHP

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Exemplo 11 (continuação):

The roots of p� s are:

s = −2

s = −1

s = 0

s = 0,5

Location of the roots of the polynomial p s

Location of the roots of the polynomial p� s

The roots of p s are:

s = −3

s = −2

s = −1

s = −0,5

Obrigado!

Felippe de Souza

[email protected]

Departamento de Engenharia Eletromecânica

Thank you!

class 10 “stability” - universidade da beira...

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